Hello, can anyone help me with these two problems?? Thank you so much in advance.

1) Prove: If x ≡ y (mod m), then (x, m) = (y, m)

2) Show that if n > 4 is not prime, then (n-1)! ≡ 0 (mod n).

January 22nd 2007, 04:45 PM

ThePerfectHacker

Quote:

Originally Posted by jenjen

Hello, can anyone help me with these two problems?? Thank you so much in advance.

1) Prove: If x ≡ y (mod m), then (x, m) = (y, m)

Let .

I presume it means .

We know that, thus, for some .

Let .
Let .
We will prove it by trichtonomy.

Assume .
But that cannot be because,
And and .
Thus, thus . And we also know that .
Thus, because . And hence it is not "greatest".

By similar reasoning. We can show leads to contradiction.

Thus, by trichtonomy,

January 22nd 2007, 05:18 PM

ThePerfectHacker

Quote:

Originally Posted by jenjen

2) Show that if n > 4 is not prime, then (n-1)! ≡ 0 (mod n).

There are two possibilities.

1)n is not a square.

2)n is a square.

If n is not a square then it has a non-trivial proper factorization where .
Where are distinct.
Thus among the factors of :
We can find its factors .

When is a square there are 2 possibilities.
1)n is not a square of a prime.
2)n is a square of a prime.

If n is not a square of a prime we know that where because it is not a prime. Thus, it has a factorization in the form where are distinct and the same argument applies.

If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that and in no other way. We know that contains one factor. But what about other another factor of ? It turns out that when the factor appears among . Thus, there is another number that has a factor of . The reason why is because for thus, and is hence among one of the factors.

January 22nd 2007, 07:31 PM

jenjen

thanks theperfecthacker for the quick reply!

February 19th 2007, 09:51 PM

Ideasman

Quote:

Originally Posted by ThePerfectHacker

There are two possibilities.

1)n is not a square.

2)n is a square.

If n is not a square then it has a non-trivial proper factorization where .
Where are distinct.
Thus among the factors of :
We can find its factors .

When is a square there are 2 possibilities.
1)n is not a square of a prime.
2)n is a square of a prime.

If n is not a square of a prime we know that where because it is not a prime. Thus, it has a factorization in the form where are distinct and the same argument applies.

If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that and in no other way. We know that contains one factor. But what about other another factor of ? It turns out that when the factor appears among . Thus, there is another number that has a factor of . The reason why is because for thus, and is hence among one of the factors.

Hmm, you have:

2 <= a,b <= n - 1, shouldn't it be: 1 < a < b < n - 1? Maybe you got it mixed up for when it is a square. The proof I was given was:

Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.

Does this work? It contradicts a few of your results.

February 20th 2007, 11:37 AM

ThePerfectHacker

Quote:

Originally Posted by Ideasman

Hmm, you have:

2 <= a,b <= n - 1, shouldn't it be: 1 < a < b < n - 1? Maybe you got it mixed up for when it is a square. The proof I was given was:

Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.

Does this work? It contradicts a few of your results.

No. I said proper nontrivial factorization. Meaning the obvious factors, 1 and the number itself are excluded.

February 21st 2007, 09:22 PM

Ideasman

Quote:

Originally Posted by ThePerfectHacker

No. I said proper nontrivial factorization. Meaning the obvious factors, 1 and the number itself are excluded.

Sorry to be a pain, but I clearly understand your proof now TPF. What I don't understand is how the proof I was given (below) proves the claim; how does that tie it all together showing that n is composite if n > 4 and that n will then divide (n - 1)!

Proof:

Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.

February 22nd 2007, 03:40 PM

ThePerfectHacker

Quote:

Originally Posted by Ideasman

Sorry to be a pain, but I clearly understand your proof now TPF. What I don't understand is how the proof I was given (below) proves the claim; how does that tie it all together showing that n is composite if n > 4 and that n will then divide (n - 1)!

Proof:

Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.

How do you know that 2a<=n-1?
And hence among the factors.
See I acutally used an inequality to justify that.

February 22nd 2007, 03:57 PM

Ideasman

Quote:

Originally Posted by ThePerfectHacker

How do you know that 2a<=n-1?
And hence among the factors.
See I acutally used an inequality to justify that.